# If f:[1,infinite)-->[2,infinite) is given by f(x)=x +1/x then f^-1(x) is equal to

Solution:   We can write    $y=\frac{x^{2}+1}{x},$

or           $x^{2}-yx+1=0$

$\Rightarrow$         $f^{-1}(y)=x=\frac{y\pm \sqrt{y^{2}-4}}{2}$

$\Rightarrow$          $f^{-1}(x)=\frac{x+\sqrt{x^{2}-4}}{2}$

Note that   $f^{-1}(x)=\frac{x-\sqrt{x^{2}-4}}{2}$  is not possible because

$f^{-1}:[2,\infty )\rightarrow [1,\infty )$    and  $f^{-1}(3)\notin [1,\infty ).$

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